# Interaction of the three kinds

There're three kinds of it.

• Any two of the same kind will give you that same kind.

• Any two of the different kinds will give you the remaining kind.

What is it?

I don't know the answer. I'm trying to find a real-world embodiment for this concept.

Since there was a lot of misunderstanding I'll try to elaborate. Formally, we have three types (classes, families, kinds) of something, denote them $$A,B$$ and $$C$$. We can take instances of those types, like $$a$$ of $$A$$ ($$a\in A$$ as a notation), where $$a$$ is some specific object that is assosiated with the class $$A$$. We also have a mapping (combining, mixing, colliding) procedure $$\times$$ that for any $$a_1,a_2\in A$$, $$b_1,b_2\in B$$, $$c_1,c_2\in C$$:

$$\qquad a_1\times a_2\in A,\quad b_1\times b_2\in B,\quad c_1\times c_2\in C$$

$$\qquad a_1\times b_1\in C,\quad b_1\times c_1\in A,\quad c_1\times a_1\in B$$

The goal is to find a natural embodiment of these types, instances and mapping procedure that behaves by these rules without us having to explicitly explain them.

We can make up a bunch of forced examples, like the mechanics of some objects in a video game, or a part of some logic mechanism or schematics. But to construct such an example, we would actually need to lay out the rules once again. They don't emerge naturally and that's the problem.

• Is there a mathematical name for this from which you were inspired to ask this question? This looks similar (but not exactly) to quaternion. – justhalf Jul 26 '16 at 8:34
• @justhalf, I wish there was, it would've been an answer to the question. Of course in mathematics we could force an operation to behave this way, but it wouldn't be natural. It would be just another formulation of the concept, not its embodiment. – Glinka Jul 26 '16 at 8:44
• The problem with the operation you want is that it can't form a Group (it is not associative and does not contain the identity), so you'll definetly not find anything in physics. – FrodCube Jul 26 '16 at 9:08
• @FrodCube, if something doesn't form a group, it's not yet to be write off. $\mathbb{R}$ under involution is not a group. Kirchhoff's circuit laws don't form a group. There are lots of different algebraic structures, or not necessarily algebraic. Why a group? – Glinka Jul 26 '16 at 9:48
• @justhalf, yes, thank you! The mathematical operation is certainly a breakthrough. And the card example also fits. Cool ;) – Glinka Jul 26 '16 at 9:54

As pointed out by FrodCube in the comment, the structure cannot be a group. The structure also cannot be a ring, since by your requirement, $(x*x)*y = x*y = z \neq y = x*z = x*(x*y)$, so the operation is not associative, which is required by ring.

A mathematical structure that satisfies the two conditions is the structure with three elements $\{0, 1, 2\}$ and the operation: $*$ such that:

$x*y = -(x+y) \mod 3$

Since:

Condition 1:
$0 * 0 = -(0 + 0) \mod 3 = 0$
$1 * 1 = -(1 + 1) \mod 3 = -2 \mod 3 = 1$
$2 * 2 = -(2 + 2) \mod 3 = -4 \mod 3 = 2$

Condition 2:
$0 * 1 = -(0 + 1) \mod 3 = -1 \mod 3 = 2$
$1 * 2 = -(1 + 2) \mod 3 = -3 \mod 3 = 0$
$2 * 0 = -(2 + 0) \mod 3 = -2 \mod 3 = 1$
Also they are commutative, so $x*y=y*x$

Note that the "elements" above can serve as categories also, for example:

Let
$A = \{\ldots, -6, -3, 0, 3, 6,\ldots\},$
$B=\{\ldots, -5, -2, 1, 4, 7, \ldots\},$
$C=\{\ldots, -4, -1, 2, 5, 8, \ldots\}$.
The operation $\times$ defined as $x\times y = -(x+y)$ will satisfy the formal requirements specified in the question.

If this is "real enough" for you as embodiment, then that's great. Otherwise we need to find "more real" example, maybe something like this:

- Let's say we have three types of cards X, Y, Z. Then we gain a combo when either we have three of the same kind or one from each kind. Then the operation you are looking for would be "the third card such that we have a combo given the first two cards". Like f(X, X) = X (to get the "three of the same kind" combo), f(X, Y) = Z (to get the "one from each kind" combo), etc.
Some commenters point out that the game Risk or Set might be some contexts where this operation can take place in real life.

For some other ideas:

- Some form of Scissor-Paper-Stone may work?
- Take some ideas from here.

• this seems to fit. I can't see how this can be found in nature tho (an answer to this puzzle) – pwnsauce Jul 26 '16 at 9:45
• @pwnsauce: Yea, it seems hard to find examples from nature. But since card games are already ubiquitous, I guess that counts as "natural"? haha – justhalf Jul 26 '16 at 9:46
• The card game example is exactly the requirements for cashing in cards in Risk, i.e. "Given that you have two Risk cards, what card type do you need to cash them in for more armies?" – theosza Jul 26 '16 at 11:48
• The last card game example also fits with Set: for each attribute (e.g. colour), there are three options, and you want either one-of-each or three-of-one. – Peter LeFanu Lumsdaine Jul 26 '16 at 14:17
• It could also be the average mod 3, because division by 2 is equivalent to negation. – f'' Jul 26 '16 at 15:30

Draw three lines intersecting at 60 degree angles:

The reflection of any line over itself is itself. The reflection of any line over any other line is the third line.

• This is cool, as the interpretation of the first and second argument is different, but they are still commutative =D. Maybe you can improve this answer by providing more explicit definition of the operator, such as "Let the operation $a\times b$ defines the line created by reflecting $a$ on $b$" – justhalf Aug 1 '16 at 2:02

I guess there is
a + b = c
a + c = b
b + c = a

Plus:
a + a = a
b + b = b
c + c = c

Which is only true for a = b = c = 0 leading to only one kind. Correct me if I'm wrong.

Real-world embodiment:
1. no time
2. a lot of money
3. a lot of work

The combinations are working from my point of view but I'm not sure if this is subjective.

• What you wrote can be considered as a more formal formulation of the concept, with some abstract operation $+$. It clearly doesn't work for the ordinary $+$. – Glinka Jul 26 '16 at 8:47
• @Glinka does it have to be + and -? Or could we also consider * and / ? Also I have kind of an embodiment that will give you the opposite of the third. I will try on ;) – Avigrail Jul 26 '16 at 8:55

This partly fits, in that you can derive C from just A and B...

Take three vectors with sum zero, A, B and C.

Taking two vectors from the three (that can be the same or not the same), and summing those two vectors, you would get:

2A from A and A
2B from B and B
2C from C and C
-A from B and C
-B from A and C
-C from A and B

So:

• Any two of the same kind will give you TWICE that same kind
• Any two of the different kinds will give you the OPPOSITE OF the remaining kind

Edit: Of course, this isn't that special for vectors, it works for any numbers! If three numbers add up to 0, you can find the third from the first two...

• Hmm, but if $A+B+C=0$, then $\frac{B+C}{2}=-\frac{A}{2}$, not $-A$. – justhalf Jul 26 '16 at 15:07
• @justhalf, you are right, of course! Will try to salvage something useful from my answer :) – Ali Jul 26 '16 at 15:40

Is it flag with colors 'blue, yellow and green'?

same kind will give you that same kind
color mixed with itself gives same color
Two of the different kinds will give you the third kind
Blue + Yellow = Green

• I don't see how this fits... Blue + Yellow = Green? And anyway, we need that any two of the different kinds give another kind – Glinka Jul 26 '16 at 8:29
• first I was gonna say states of matter...solid+solid;liquid+liquid;but then it comes solid+liquid and I am stuck :( – smriti Jul 26 '16 at 8:46

Partial fit -

It can be rectangular coordinates (x,y,z) and the operation be vector product.

two of the same kind will give you that same kind

x X x=y Xy=z X z= 0.

two of the different kinds will give you the remaining kind

x X y=z ,y X z=x ,zXx =y

• 0 is not the same kind as $x$. OP expects $x*x = x$. – justhalf Jul 26 '16 at 9:18

Not a real world example, but taking
Group A: vectors of form $(x,0,0)$
Group B: vectors of form $(0,x,0)$
Group C: vectors of form $(0,0,x)$

and Vector multiplication (the cross product) as the "interaction".

Edit: Cross product gives a vector perpendicular to both the operands, so $a \times b \in C$
but after rechecking, $a \times a = 0$... which makes this answer wrong.

• Would it be possible for you to include a detailed explanation as to why you believe this answer is correct? Thank you! – user20 Jul 26 '16 at 14:48
• I've edited your post to use MathJax (our math rendering software) and cleaned up the grammar. Let me know if I've accidentally changed the intention of the post! – Deusovi Jul 26 '16 at 20:38

I won't hide my answer because you are actually looking for a solution. Others may see this and come up with something better.

My answer is based upon the Ideal Gas Law:

$$PV=nRT$$

Where $P = Pressure$, $V = Volume$, and $T=Temperature$. Assume the others are constants.

Let $A=\Delta{T}$, then $\Delta{T}$ plus another $\Delta{T}$ produces an overall $\Delta{T}$. $A+A=A$

Let $B=\Delta{V}$, then $\Delta{V}$ plus another $\Delta{V}$ produces an overall $\Delta{V}$. $B+B=B$

Let $C=\Delta{P}$, then $\Delta{P}$ plus another $\Delta{P}$ produces an overall $\Delta{P}$. $C+C=C$

This produces the following based on the Ideal Gas Law equation:

$\Delta{T}$ + $\Delta{V}$ produces $\Delta{P}$ based on $P=\frac{nRT}{V}$. $A + B=C$

$\Delta{T}$ + $\Delta{P}$ produces $\Delta{V}$ based on $V=\frac{nRT}{P}$. $A + C=B$

$\Delta{V}$ + $\Delta{P}$ produces $\Delta{T}$ based on $T=\frac{PV}{nR}$. $B + C=A$

In other words, "It" would be "Change". Like changes result in a like overall change. Different changes result in yet another change.

This is the best I can imagine right now. Maybe someone can find another triple-variable equation.

Two points in a finite projective plane determine a line. If the finite projective plane is taken to be over the field of three elements, then each coordinate of the operation that takes two points and produces the third point that is colliear with the two points has the desired property.